cos(x+30)cos(x+45)cos(x+105)+(cosx)^3=0,求 tanx

2cos(x+105)cos(x+45)
=cos(2x+150)+cos(60)
=cos(2x+150)+1/2

-2(cosx)^3
=2cos(x+30)cos(x+45)cos(x+105)
=cos(x+30)[cos(2x+150)+1/2]
=(1/2)cos(x+30)+(1/2)2cos(x+30)cos(2x+150)
=(1/2)cos(x+30)+(1/2)[cos(3x+180)+cos(x+120)]
=(1/2)cos(x+30)-(1/2)cos(3x)+(1/2)cos(x+120)

-4(cosx)^3=cos(x+30)-cos(3x)+cos(x+120)=cosxcos30-sinxsin30+cosxcos120-sinxsin120-cos(3x)=cosxcos30-sinxsin30+cosxcos120-sinxsin120+3cosx-4(cosx)^3

cosxcos30-sinxsin30+cosxcos120-sinxsin120+3cosx=0
cos30-tanxsin30+cos120-tanxsin120+3=0
tanx(sin30+sin120)=3+cos30+cos120
tanx=(3+√3/2－1/2)/(1/2+√3/2)